Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7
might become 4
5 6 7 0 1 2
).
Find the minimum element.
You may assume no duplicate exists in the array.
排好序的数列是很容易搞定的,直接找最小值
class Solution {
public:
int findMin(vector<int> &num) {
int min=num[0];
int i;
for(i=0;i<num.size();i++)
{
if(num[i]<min)
{
min=num[i];
break;
}
}
return min;
}
};
二分查找,经典的查找算法
class Solution {
public:
int findMin(vector<int> &num) {
int l,r,min,mid,size;
size = num.size()-1;
min = num[0];
l=0;
r=size;
while(l<r)
{
if(num[l]<num[r])
{
break;
}
else
{
mid=(l+r)/2;
if(num[mid]<num[l])
{
r=mid;
}
else
{
l=mid+1;
}
}
}
return num[l];
}
};